R1= K1 A1ℓ1,R2= K2 A2ℓ2dtdQ=RΔT(dtdQ)1=(μdQ)2R1400−T=R2T−200T−200400−T=R2R1=(ℓ2ℓ1)(r1r2)2× K1K2=21×(21)2×2=(41)T−200400−T=411600−4 T=T−2005 T=1800 T=360 K
JEE Main 2025 — Physics Thermodynamics
Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are :
LBLA=21,rBrA=2 and KBKA=21. The free ends of rods A and B are maintained at 400 K,200 K, respectively. The temperature of rods interface is ________ K, when equilibrium is established.
Held on 7 Apr 2025 · Verified 6 Jul 2026.
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Work through every JEE Main Thermodynamics PYQ, year by year.